Question

find the first four consecutive terms in A.P. whose sum is 12 and the sum of 3rd and 4th term is 14.

Answer 1

hello Frnd

Given as

Sum of First for term of an AP = 12
And, also sum of third term and 4th term is = 14

As we know that

$s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ here \: s_{4} \: is \: given \: as \: 12 \\ \\ \\ \\ s_{4} = \frac{4}{2} (2a + 3d) \\ \\ 12= \frac{4}{2} (2a + 3d) \\ \\ 2a + 3d = 6................(1)$

Now,

It is also given that

$a_{3} + a_{4} = 14 \\ \\ a + 2d + a + 3d = 14 \\ 2a + 5d = 14.............(2)$

Now,

(2) - (1); We get

2a -2a + 5d - 3d = 14 - 6
2d = 8
d = 4

When d = 4 putting in (1)

2a + 3( 4) = 6
2a = 6 - 12
2a = -6
a = -3


Hence the required AP is

-3, 1, 5, 9, 13 ......



hope it helps
jerri

Answer 2

hello



Let the four consecutive terms in an A.P. be a – 3d, a – d, a + d, a + 3d


As per the first condition,


a – 3d + a – d + a + d + a + 3d = 12


∴  4a = 12


∴ a = 12/4


∴ a = 3 ...........eq. (i)


As per the second condition,


a + d + a + 3d = 14


∴ 2a + 4d = 14



∴ 2 (3) + 4d = 14 [From eq. (i)]


∴ 6 + 4d = 14


∴ 4d = 14 – 6


∴ 4d = 8


∴ d = 8/4


∴ d = 2


∴ a – 3d = 3 – 3 (2) = 3 – 6 = – 3


a – d = 3 – 2 = 1


a + d = 3 + 2 = 5


a + 3d = 3 + 3 (2) = 9

∴ The four consecutive terms of A.P. are – 3, 1, 5 and 9.